Coding and maths classes in Berlin

Source: IQB, Abitur pages and Beschreibung der Struktur der Aufgaben (Mathematik), read 2 November 2026. Since 2017 the pools cover German, maths, English and French; since 2022 those exams have been held largely on common dates. Each Land, Berlin included, decides through its own commission which pool tasks it uses and how far it adds its own. The Senate's 2027 exam plan could not be fetched when we checked (the server asked us to wait), so this page prints no Berlin exam date.

import csv

with open("breitband_2025.csv", encoding="utf-8") as f:
    schools = list(csv.DictReader(f, delimiter=";"))

by_status = sum(1 for s in schools if s["aktueller status"] == "Anschluss nutzbar")
by_flag = sum(1 for s in schools if s["anschluss nutzbar"] == "Ja")
print("Schools:", len(schools))
print("Status says usable:", by_status, round(100 * by_status / len(schools), 1), "%")
print("Yes/no column says usable:", by_flag, round(100 * by_flag / len(schools), 1), "%")

Two columns, two answers: 594 by the status, 590 by the yes or no field. The difference is small, but a report that quotes 83.3 % in one paragraph and 82.7 % in the next loses trust. Version 2 lists the rows behind the gap.

clash = [s["schulnummer"] for s in schools
         if (s["aktueller status"] == "Anschluss nutzbar") != (s["anschluss nutzbar"] == "Ja")]
print("Rows where the two columns disagree:", clash)

Four schools are marked usable in the status column and Nein in the final column. In all four, the civil works and the equipment installation are dated 2025, which suggests the yes or no field had not caught up when the extract was made, but the file does not say, so we report both counts and name the four rows rather than guess. That is the habit: when two fields should agree and do not, show the disagreement.

from datetime import datetime
from statistics import mean, median

def day(text):
    try:
        return datetime.strptime(text.strip(), "%d.%m.%Y").date()
    except ValueError:
        return None

waits = []
for s in schools:
    visit, kit = day(s["begehung "]), day(s["aktive technik verbaut "])
    if visit and kit:
        waits.append((kit - visit).days)
print("Schools with both dates:", len(waits))
print("Mean wait:", round(mean(waits), 1), "days; median:", median(waits), "days")
print("Shortest:", min(waits), "Longest:", max(waits))

Data: Senatsverwaltung für Bildung, Jugend und Familie, Breitband-Ausbau der Berliner Schulen (November 2025 extract), licence CC BY, downloaded 3 November 2026. Counts, percentages, means and medians are our own calculations; the file gives no reasons for differences between rows or districts, and neither do we.

Here the mean (529.8 days) and the median (537 days) are close, so the typical school waited about a year and a half, and neither number misleads. The maths lesson is to check that before choosing one: had a handful of schools waited five years, the mean would have jumped while the median barely moved. The wait of 0 days at one site is worth a second look too, since a visit and an installation on the same day is unusual.

1. Klasse 7 and 8, by hand. Five friends wait 3, 4, 4, 5 and 30 minutes for a bus. Work out the mean and the median. Which one describes a normal wait?
2. Klasse 9 and 10, Python. Run Version 1 and explain in one sentence why the two percentages differ. Then write Version 2 before reading ours.
3. Oberstufe, statistics. Compute the quartiles of the waiting times and draw a box plot. Is any district an outlier by the 1.5 times IQR rule?
4. Adults. Turn Version 3 into a function that returns mean, median and the share of waits over two years, then test it on a list you make up.